Remark:L’Hˆopital’s rule is not allowed in Math 1225; it’s first introduced in Math 1226.Only use methodscovered in class. In particular, L’Hˆopital’s rule requires derivatives - a topic we have yet to cover in this class.Remark:Functions that look like00at a particular point need to be manipulated further.Limits are deter-mined by the behavior of the function around a pointx=a, not the function value atx=a.limx→af(x) may exist even thoughf(a) is undefined.Limits involving an algebraic combination of functions in an independent variable may often be evaluated byreplacing these functions by their limits; if the expression obtained after this substitution does not give enoughinformation to determine the original limit, it is said to take on anindeterminate form. Typical indeterminateforms look like00,∞∞,0· ∞,1∞,∞ - ∞,00,∞0Such limits need to be manipulated further in order to be evaluated.Example:Consider the graphs offandgEvaluatelimx→-2[f(x) + 5g(x)],limx→1[f(x)g(x)],and limx→2f(x)g(x).First we see that limx→-2f(x) = 1 and limx→-2g(x) =-1. Thereforelimx→-2[f(x) + 5g(x)]=limx→-2f(x) +limx→-2[5g(x)] by the sum law=limx→-2f(x) + 5 limx→-2g(x) by the constant multiple law=1 + 5·(-1) =-4Next we see that limx→1f(x) = 2, however limx→1g(x) DNE since limx→1-g(x) =-26= limx→1+g(x) =-1. Sowe can’t use the product law as the hypotheses are not all satisfied. However we can use it for the one-sided limitslimx→1-[f(x)g(x)] =limx→1-f(x)·limx→1-g(x)= 2·(-2) =-4limx→1+[f(x)g(x)] =limx→1+f(x)·limx→1+g(x)= 2·(-1) =-2Since limx→1-[f(x)g(x)]6= limx→1+[f(x)g(x)], we can now deduce that limx→1[f(x)g(x)] DNE.Finally we see that limx→2f(x)≈1.4 and limx→2g(x) = 0. We cannot use the quotient law since the denomi-nator vanishes. However since the numerator doesn’t vanish we can conclude that limx→2f(x)g(x)DNE.Example:Considerf(x) =x2-1x-1andg(x) =x2-3x+ 2x-1

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Try to computef(1) andg(1). Determine limx→1f(x) and limx→1g(x) using the graphs of each function. Is itaccurate to say thatf(x) =x+ 1?Sincefandgare undefined atx= 1 we cannot computef(1) org(1). From the graphswe find that limx→1f(x) = 2 and limx→1g(x) =-1.Finally we must be precise with limit notation. It is correct to writelimx→1f(x) = limx→1(x+ 1),whereasf(x) =x+ 1 is not.f(x) is not defined atx= 1, whereasy=x+ 1 is defined atx= 1. Away fromx= 1, these functions are equal;limits are calculated near the point of approach, so we may cancel holes in the presence of limit notation.Example:Evaluatelimt→-3t2-92t2+ 7t+ 3First note that att=-3 the expression looks like00, hence we need to manipulate it further. Via factorization weobtainlimt→-3t2-92t2+ 7t+ 3= limt→-3(t+ 3)(t-3)(2t+ 1)(t+ 3)= limt→-3t-32t+ 1=65by the sum, difference, constant multiple, and quotient laws.Example:Evaluatelimx→-4√x+ 29-5√20 +x-4First check that att=-4 the expression looks like00, so we manipulate it further (if it was a well defined finitenumber then only direct substitution and the limit laws would have been required). In this example factorization